## What, if anything, makes homogeneous polynomials so great?

It should be obvious from the question that I am not any kind of algebraic geometer, so if there are definitions of hom-polys as comonoidal dyadic functors or whatnot, let's leave that to one side for the purposes of this question. I really mean hom-polys in the most pedestrian sense possible.

From the outside, it seems that homogeneous polynomials get a lot of attention in alg-geom. (Perhaps in other areas as well?) I know that they are, well, homogeneous with respect to dilations, and that this allows one to look at their zeros in projective space in a natural way.

I usually like to keep a safe distance between myself and projective space, and have always looked at hom-polys as "merely" a technical tool. But, just the other day, I was able to quickly solve a small, elementary number-theoretic problem by converting it into "homogeneous" form. (Some vague memories of homogeneous problem-solving heuristics prompted this.) To be precise, the problem was that of computing how many solutions there are to m^2 + n^2 = 1 in Z_p, where p is a prime congruent to 3 mod 4. (Of course this is trivial by a change of variables when p = 1 mod 4.) The problem seemed unfriendly at first, but passing to the related question of the solutions to m^2 + n^2 - t^2 = 0 revealed a lot of symmetry and made it quite trivial. (Of course, solving small cases of the first problem showed a lot of symmetry, but it wasn't obvious how to get a handle on it.)

My question is: do hom-polys help to solve a lot of seemingly unrelated problems via some process of homogeneization of the problem? Do you have examples? Is this one reason for their popularity? I'm looking for heuristics mostly, but if you think an actual theorem makes precise or helps formulate a heuristic, go nuts.

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This is a bit too short for an answer, so I'll just leave it here: What makes the projective plane so great? The fact that it doesn't have many of the annoying limitations that Euclidean geometry has. In Euclidean geometry, two lines may intersect at one point, but in some rare but still existent cases will be parallel, and thus you have two cases to consider - which in fact means $2^n$ cases if you have more than just two lines. In projective geometry, the second case is almost nonexistent - the only bad thing that can happen is that the two lines coincide, which is much more seldom than... – darij grinberg Jun 1 2010 at 21:03
... parallelism. So the two cases are reduced to one (well, except of the coinciding case, but as I said, it is much more seldom and easier to rule out usually). That makes projective geometry easier to handle than Euclidean/affine geometry (when I say Euclidean/affine, I mean $\mathbb R^2$ as opposed to $\mathbb P\mathbb R^2$; I am not talking about distances, angles etc.). Now, the functions of interest on the Euclidean plane are polynomials, while those on the projective plane are homogeneous polynomials... – darij grinberg Jun 1 2010 at 21:05
Well, fractions of homogeneous polynomials actually, but that doesn't make things more difficult. – darij grinberg Jun 1 2010 at 21:05
Another reason homogeneous polynomials are popular in introductory discussions of algebraic geometry is that they are stand-ins for global sections of line bundles. If $X$ is a projective scheme, and $L$ a very ample line bundle on $X$, then $L$ gives an embedding into some $P^n$. By Serre vanishing, for large enough powers of $d$, $H^{0}(X,L^d)$ can be identified with the quotient of the degree $d$ homogeneous polynomials (on $P^n$) by the ideal of $X$ in degree $d$. This equivalence allows us to talk around global sections of line bundles without actually introducing line bundles.... – Mike Roth Jun 1 2010 at 21:16
Another interesting fact is this: in projective space, varieties of complementary dimensions always intersect. Thus, it is much easier to find (in the nonconstructive sense) nontrivial solutions to homogeneous systems of polynomials, than solutions to inhomogeneous systems. (The portion of Darij's remark above about parallel lines is a special case of this.) – Charles Staats Jun 1 2010 at 21:35
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I would rephrase the question as "what is so great about projective space (as compared to affine space)?" I would give two answers:

1. Projective space has a larger symmetry group: dimension $n^2+2n$ rather than $n^2+n$.
2. Projective space is compact.

The first is what you are using when you turn $x^2+y^2=1$ into $x^2+y^2=z^2$ and then $(z+y)(z-y)=1$; the corresponding change of coordinates is a symmetry of projective space which doesn't pass to affine space. The second is why homogenous polynomials work better for intersection theory: intersections can't run off to infinity. It is also why point counting over finite fields gives nicer answers in projective space: point counting is related to cohomology, and cohomology for smooth compact spaces obeys Poincare duality.

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I gather that there is also some deeper sense in which projective varieties are somehow tamer than other complete varieties. I could make a stab at a summary of the reasons, but I might be interested in an insider's explanation of why (or whether) it is so. – Greg Kuperberg Jun 2 2010 at 6:11
Hi David, thanks for your answer! Phrasing the success of the construction as a "change of coordinates which doesn't pass to affine space" is a very intuitive and elegant way of putting it. That was very helpful. I'm familiar with the "running off to infinity" problem in analysis, but I guess I'll have to think about it more in the arithmetic setting. Thanks again! – Pietro KC Jun 3 2010 at 6:08
Greg, if you can find the time, I would be delighted to hear what you have to say on tameness of varieties! – Pietro KC Jun 3 2010 at 6:09

From a practical perspective, putting a grading on an algebra usually organizes the algebra into a collection of finite-dimensional vector spaces, each indexed by a natural number. This opens the door to induction arguments which at each step, only have to deal with a finite-dimensional vector space. Its this crude idea which seems to motivate most of the computational techniques in the theory of commutative algebras (see, anything with Gr\"obner bases).

More generally, graded algebras/projective spaces allow finite-dimensional-type techniques to be used in the study of infinite-dimensional algebras and modules. As an example, if $A$ is a polynomial ring, and $M$ is a f.g. graded $A$ module, then the double dual $$Hom_\mathbb{C}(Hom_{\mathbb{C}}(M,\mathbb{C}),\mathbb{C})$$ is a monstrosity: infinitely-generated and non-graded. However, its graded double dual $$\underline{Hom}\mathbb{C} (\underline{Hom}\mathbb{C} (M,\mathbb{C}),\mathbb{C})$$ is isomorphic to $M$.

This finite-dimensionality mantra is even more prominent in the study of projective schemes. A coherent sheaf of modules $\mathcal{M}$ on $\mathbb{P}^n$ will not only have finite-dimensional global sections, but all higher cohomologies of $\mathcal{M}$ will be finite dimensional. This means you can talk about things like the dimension of these cohomologies, which allows definitions of things like 'genus' and 'Euler characteristic'. These concepts have to be heavily modified to make any sense in the affine cases.

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 Hi Greg, thanks for your answer! I was aware of the inductive uses of grading an algebra; due to my background I guess the example that struck me the most was Dvir's proof of the finite field Kakeya conjecture. I don't fully appreciate your last paragraph, since "coherent sheaf of modules on P^n" and "global sections" are not things on which I have much intuition. The most familiar interpretation for me would be the sheaf on P^n which assigns to each open set U the $\mathbb{R}$-module of continuous (real) functions on U. Would a global section just be a continuous function on all of P^n? – Pietro KC Jun 3 2010 at 5:37 I don't understand what finite-dimensionality of global sections would be. If P^n is real projective space it seems like the global continuous functions would be an infinite-dimensional space, so I guess I have the wrong idea. Please forgive the elementary questions! If this would take too long to explain then just a reference would be great. – Pietro KC Jun 3 2010 at 5:46 I'm thinking about complex projective space when I say P^n. Here, you want the sheaf of algebraic functions. Since algebraic functions are always complex analytic, they satisfy the maximum modulus principle, which states that they attain their maximum norm on the boundary of the domain they are defined on. The function must then be constant, since for any point p, the norm of the function at p must be greater than or equal to the norm on all of P^n\p. – Greg Muller Jun 3 2010 at 15:57 The reason for algebraic functions and not continuous ones can be traced back to the kind of ring we care about. the scheme P^n is closely related to C^{n+1}, and the ring of algebraic functions on C^{n+1} is C[x_0,...x_n], while the ring of continuous functions is horrible from an algebraic perspective (infinitely-generated, non-Noetherian, etc). In general in algebraic geometry, you want algebraically closed fields and algebraic functions, unless you have a very good reason otherwise. – Greg Muller Jun 3 2010 at 15:59

Well, if you are interested in counting solutions mod p, you could note that the "good" formulae are indeed related to the homogeneous approach/projective space. It is not just a question of restoring the symmetry between variables, though that can be part of it: there are more projective transformations than affine transformations. The general theory of local-zeta functions would explain that counting points mod p works best in the homogeneous setting, and making things inhomogeneous is going to cut out some points, in a way that is relatively random.

Try http://en.wikipedia.org/wiki/Pl%C3%BCcker_coordinates for example, for geometry.

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 Your remark about the random cutting out of points is very helpful, thanks! I hadn't looked at it this way. Coincidentally, just today I listened to a talk on local zeta functions and it was also very clarifying. – Pietro KC Jun 3 2010 at 5:20

One thing you can do with them is to solve the following problem: given $f_1,\ldots,f_k$ polynomials in $x_1,\ldots,x_n$, is there a polynomial over a field $K$, call if $g$, such that $g=0$ if and only if $f_1=\ldots=f_k=0$? The solution that I know of involves taking an irreducible polynomial over $K$ (so $K$ can't be algebraically closed), homogenizing it, and then substituting the original polynomial in for the new variable, and iterating the process. The solution may not be homogeneous, but homogeneous polynomials are useful for solving some problems stated entirely in terms of arbitrary polynomials.

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 Hi Charles, thanks for your answer! It seems like your remark is spot-on (using hom-polys to solve another problem), but I don't quite understand the construction. Are the f_i given over K, and then you want to find a g, also over K, the zeros of which are exactly the intersection of the zeros of all f_i? Or are the f_i given over some F and you can then choose your K? Even granting the 1st option, I'm not sure I understand the construction: what is the "original polynomial"? How are the f_i coming in? – Pietro KC Jun 3 2010 at 6:02 There's only one field, $K$. Take $f(x)$ to be an irreducible poly in one var over $K$, then take $f'(x,y)$ the homogenization. Then look at $f'(x,f(y))$ and iterate this construction, because $f'(x,y)=0$ if and only if $x=y=0$ in the field, because it was irreducible. – Charles Siegel Jun 3 2010 at 12:04